As I was walking through campus today, I had an interesting question pop into my head: How can we prove that the period of $\tan(x)$ is $\pi$ rather than $2\pi$? The answer to this was extremely straightforward: We start off with $$\tan(x) = \tan(x + T) = {\tan(x) + \tan(T) \over 1 – \tan(x) \tan(T)}$$ to give us $$-\tan^2(x)\tan(T) = \tan(T)$$ $$0 = \tan(T) + \tan^2(x)\tan(T)$$ $$0 = \tan(T)[1 + \tan^2(x)]$$ $$\implies \tan(T) = 0\;\;\;\;\;\text{and}\;\;\;\;1 + \tan^2(x) = 0 \implies \text{No real solution for any $x\in\mathbb{R}$}$$ Which for $\tan(T) = 0 \implies {\sin(T) \over \cos(T)} = 0 \implies \sin(T) = 0$, we have $T = 0, \pi \implies T = \pi$ to show that the period of $\tan(x)$ is $\pi$ if we desire a nontrivial answer.
But I got stuck trying to do the same with $\sin(x)$. I tried:
$$\sin(x) = \sin(x + T) = \sin(x)\cos(T) + \sin(T)\cos(x)$$ $$\implies \sin(x)[1 – \cos(T)] = \sin(T)\cos(x)$$ $$\implies \tan(x) = {\sin(T) \over 1 – \cos(T)}$$
But I got stuck here. I'm not sure how to isolate a single trig function in terms of $T$.
I Googled this proof, but everyone either uses Taylor Expansions, Euler's Formula, or calculus. But I'm looking for an argument I could present to someone with knowledge of trigonometry and no more. Any ideas?
Best Answer
If $T$ is such that for all $x\in \mathbb{R} $ we have $\sin(x+T) =\sin(x) $, then in particular, setting $x=0$, we have
$$\sin T =0$$
So $T=k\pi$ with $k \in \mathbb{Z} $. To conclude, we then need to check that $\sin(x+2\pi) =\sin(x)$ using your formula above (and that $\pi$ is not a period, by plugging $x=-\pi/2$ for example).